An electric kettle takes 4 ~A current at 220 ~V . How much time will it take to boil 1 ~kg of water from temperature 20^ C ? The temperature of boiling water is 100^ C

Correct answer: A. 6.3 ~min Key Idea : Heat eyolved due to Joule's effect is used up in boiling water. As per key idea aligned V I t & =m s Δ t or t & =(m s Δ t)/(V) aligned Putting under given values gathered I=4 A, V=220 volt , m=1 kg, Δ t=(100-20)^ C, s=4200 J / kg^ C t=(1 × 4200 × 80)/(220 × 4)=6.3 min gathered

NEET 2008

Exam: NEET
Year: 2008
Subject: Physics
Chapter: Thermal Properties of Matter

NEET 2008 · Physics · Thermal Properties of Matter

An electric kettle takes 4 ~A current at 220 ~V . How much time...

An electric kettle takes

4 \mathrm{~A}

current at

220 \mathrm{~V}

. How much time will it take to boil

1 \mathrm{~kg}

of water from temperature

20^{\circ} \mathrm{C}

? The temperature of boiling water is

100^{\circ} \mathrm{C}

Options

A $6.3 \mathrm{~min}$
B $8.4 \mathrm{~min}$
C $12.6 \mathrm{~min}$
D $4.2 \mathrm{~min}$

Correct answer

A. $6.3 \mathrm{~min}$

Step-by-step solution

Key Idea : Heat eyolved due to Joule's effect is used up in boiling water. As per key idea

\begin{aligned} V I t & =m s \Delta t \\ \text{or } t & =\frac{m s \Delta t}{V} \end{aligned}

Putting under given values

\begin{gathered} I=4 \mathrm{ A}, V=220 \text{ volt }, \quad m=1 \mathrm{ kg}, \\ \Delta t=(100-20)^{\circ} \mathrm{C}, s=4200 \mathrm{ J} / \mathrm{kg}^{\circ} \mathrm{C} \\ \therefore \quad t=\frac{1 \times 4200 \times 80}{220 \times 4}=6.3 \mathrm{ min} \end{gathered}

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